# Giant Swing, Uniform Circular Motion

[SOLVED] Giant Swing, Uniform Circular Motion

## Homework Statement

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.00 long, the upper end of the cable being fastened to the arm at a point 3.00 from the central shaft.

A) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of with the vertical.

B) Does the angle depend on the weight of the passenger for a given rate of revolution?

## Homework Equations

R = Lsin(\theta)

v = $$\sqrt{gtan(\theta)R}$$

T = 2$$\Pi$$R/v

## The Attempt at a Solution

tried using L = 3+5m*sin($$\theta$$)) to get 4m

R then equals = 2m

v then equals = $$\sqrt{9.8*2*tan\theta}$$ = 11.3m/s

T then equals 2$$\Pi$$(2m)/11.3m/s = 1.1s wrong

I tried a few other combinations where I used 3+(5sintheta) as the L and got T=4.4s which was wrong as well.

I'm not sure what I'm doing wrong, I think I have the correct equations and I know I have the right angle and distances, I guess I'm just not sure how to derive the right length and and radius.

I'm also thinking that for part B the weight will determine the angle that the seat swings, but I don't want to risk losing the only chance I have on that part of the problem (masteringphysics).

any and all help is greatly appreciated.

Last edited:

## Answers and Replies

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You might want to go back and look at this bit again:
L = 3+5m*sin($$\theta$$)) to get 4m

R then equals = 2m
sin(30) = 0.5, so 5*0.5 + 3 = 5.5m, and you don't need to divide it by 2, because that is the radius, not the diameter.

Clay Marisa
PhanthomJay
Science Advisor
Homework Helper
Gold Member
It appears that you are confusing L with R. If L is the length of the cable, then what is R, which is the radius measured from the passenger to the central shaft? Also, if you draw a free body diagram and use Newton's laws, it will help to see whether or not the the mass comes into play. You shouldn't blindly be using a formula..

It appears that you are confusing L with R. If L is the length of the cable, then what is R, which is the radius measured from the passenger to the central shaft? Also, if you draw a free body diagram and use Newton's laws, it will help to see whether or not the the mass comes into play. You shouldn't blindly be using a formula..
I wrote it before, R = Lsin(30)

I was pretty sure it was the length of the cable alone (which would've been 5) I just wasn't sure where the distance of the angle of the seat from the vertical shaft (3m) played in.

if it's only length of cable alone, then R = 5(sin30) so R = 2.5m, does that seem more correct?

on the angle's dependance on the weight, I've drawn a free body diagram and it makes the sum of the forces be

(sigma)Fx = -marad + mgcos30 = 0

(sigma)Fy = T - mgsin30 = 0

so I would imagine the angle is dependant on the weight of the seat.

if it's only length of cable alone, then R = 5(sin30) so R = 2.5m, does that seem more correct?
As I already said:
You might want to go back and look at this bit again:
L = 3+5m*sin($$\theta$$)) to get 4m

R then equals = 2m
sin(30) = 0.5, so 5*0.5 + 3 = 5.5m, and you don't need to divide it by 2, because that is the radius, not the diameter.

thanks, I got the answer to the first part using that radius, I wasn't picturing in my head that the swing literally makes a circle when viewed overhead and the distance of 3m + (5sin30) was the radius of that circle.

now fro for the 2nd part, as I said I did a free body diagram and found the forces acting in each direction, but I'm actually still not certain, I believe the mass would cancel out from the 2 equations if I set them equal to each other, so the answer would be that it doesn't depend on the mass?

PhanthomJay
Science Advisor
Homework Helper
Gold Member
thanks, I got the answer to the first part using that radius, I wasn't picturing in my head that the swing literally makes a circle when viewed overhead and the distance of 3m + (5sin30) was the radius of that circle.

now fro for the 2nd part, as I said I did a free body diagram and found the forces acting in each direction, but I'm actually still not certain, I believe the mass would cancel out from the 2 equations if I set them equal to each other, so the answer would be that it doesn't depend on the mass?
Your FBD's are not correct. In the y direction, mg acts straight down, and Tcos30 acts straight up. In the x direction, only Tsin30 acts.

Your FBD's are not correct. In the y direction, mg acts straight down, and Tcos30 acts straight up. In the x direction, only Tsin30 acts.
so then in that case Tcos30 = mg

so the angle is proportional to the mass?

scracth that the angle is not proportionate to the mass, thanks for the help.

I have the same problem as well, and I don't quite understand how velocity was derived.